A - Tiny Arithmetic Sequence

水题，判断3个数是否能构成等差数列

  1. void solve() {
  2.     int a, b, c;
  3.     cin >> a >> b >> c;
  4.     if (a + b == 2 * c || a + c == 2 * b
  5.             || b + c == 2 * a)cout << "Yes\n";
  6.     else cout << "No\n";
  7. }

B - Do you know the second highest mountain?

排序，先按山脉高度排序，高度一样则按名字排序

  1. struct node {
  2.     string s; int t;
  3. };
  4. vector<node>v;
  5. bool cmp(node a, node b) {
  6.     if (a.t == b.t)return a.s > b.s;
  7.     return a.t > b.t;
  8. }
  9. void solve() {
  10.     int n;
  11.     cin >> n;
  12.     for (int i = 0; i < n; ++i) {
  13.         string s; int t;
  14.         cin >> s >> t;
  15.         v.push_back({s, t});
  16.     }
  17.     sort(v.begin(), v.end(), cmp);
  18.     cout << v[1].s << '\n';
  19. }

赛后发现自己写复杂化了

  1. pair<int, string>a[1010];
  2. void solve() {
  3.     int n;
  4.     cin >> n;
  5.     for (int i = 1; i <= n; ++i)cin >> a[i].second >> a[i].first;
  6.     sort(a + 1, a + 1 + n);
  7.     cout << a[n - 1].second << endl;
  8. }

C - Secret Number

这道题，是检测 0000 ~ 9999 的每一个值，但我们可以通过高桥的字符串进行简化

  1. void solve() {
  2.     string s; cin >> s;
  3.     int ans = 0;
  4.     for (int i = 0; i <= 9999; ++i) {
  5.         vector<bool> f(10);
  6.         int x = i;
  7.         for (int j = 0; j < 4; ++j) {
  8.             f[x % 10] = true;
  9.             x /= 10;
  10.         }
  11.         bool f2 = true;
  12.         for (int j = 0; j < 10; ++j) {
  13.             if (s[j] == 'o' and !f[j]) f2 = false;
  14.             if (s[j] == 'x' and f[j]) f2 = false;
  15.         }
  16.         ans += f2;
  17.     }
  18.     cout << ans << '\n';
  19. }
  1. S = input()
  2. ans = 0
  3. for i in range(10000):
  4.     flag = [False]*10
  5.     now = i
  6.     for j in range(4):
  7.         flag[now%10] = True
  8.         now //= 10
  9.     flag2 = True
  10.     for j in range(10):
  11.         if S[j] == 'o' and not flag[j]:
  12.             flag2 = False
  13.         if S[j] == 'x' and flag[j]:
  14.             flag2 = False
  15.     ans += flag2
  16. print(ans)

D - Game in Momotetsu World

虽然正向搜索会很麻烦，但反过来从终点搜索起点使用DP记录即可

  1. using ll = long long;
  2. ll n, m, dp[2010][2010];
  3. char s[2011][2011];
  4. void solve() {
  5.     scanf("%lld%lld", &n, &m);
  6.     for (int i = 1; i <= n; i++)
  7.         scanf("%s", s[i] + 1);
  8.     memset(dp, 63, sizeof(dp));
  9.     dp[n][m] = 0;
  10.     for (int i = n; i > 0; i--)
  11.         for (int j = m; j > 0; j--) {
  12.             if (i == n && j == m)
  13.                 continue;
  14.             dp[i][j] = max((s[i + 1][j] == '+' ? 1 : -1) - dp[i + 1][j],
  15.                            (s[i][j + 1] == '+' ? 1 : -1) - dp[i][j + 1]);
  16.         }
  17.     if (dp[1][1] > 0)cout << "Takahashi\n";
  18.     else if (dp[1][1] == 0)cout << "Draw\n";
  19.     else cout << "Aoki\n";
  20. }

E - Xor Distances

  1. #include<bits/stdc++.h>
  2. #define int long long
  3. #define N 200005
  4. #define MOD 1000000007
  5. using namespace std;
  6. int n, d[N], ans;
  7. vector<int> to[N], w[N];
  8. void add(int u, int v, int wt) {to[u].push_back(v), w[u].push_back(wt);}
  9. void dfs(int u, int fa) {
  10.     for (int i = 0, v; i < to[u].size(); i++)
  11.         if ((v = to[u][i]) != fa)d[v] = d[u] ^ w[u][i], dfs(v, u);
  12. }
  13. signed main() {
  14.     cin >> n;
  15.     for (int i = 1, u, v, wt; i < n;
  16.             i++)scanf("%lld%lld%lld", &u, &v, &wt), add(u, v, wt), add(v, u, wt);
  17.     dfs(1, 0);
  18.     for (int k = 0; k < 60; k++) {
  19.         int a = 0;
  20.         for (int i = 1; i <= n; i++)a += ((d[i] >> k) & 1);
  21.         ans = (ans + a * (n - a) % MOD * ((1ll << k) % MOD) % MOD) % MOD;
  22.     }
  23.     cout << ans << endl;
  24.     return 0;
  25. }

F - Insertion Sort

待补